Problem: Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

Solution (in JAVA):

Solution:
// IT'S JAVA SO DON'T MIND THE LONG CODE :)
import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class nauuoChess {

    static void MainSolution() {
        ArrayList<two> ar = new ArrayList<>();
        for (int i = 1; i <= 500; i++) {
            ar.add(new two(i, i));
            ar.add(new two(i, i + 1));
        }
        n = ni();
        pl(ar.get(n - 1).b);
        for (int i = 0; i < n; i++) pl(ar.get(i).a + " " + ar.get(i).b);
    }
    static class two{
        int a, b;
        two(int a, int b){
            this.a=a;this.b=b;
        }
    }

    /*----------------------------------------------------------------------------------------------------------------*/
    //THE DON'T CARE ZONE BEGINS HERE...\\
    static int mod9 = 1_000_000_007;
    static int n, m, l, k, t;
    static AwesomeInput input = new AwesomeInput(System.in);
    static PrintWriter pw = new PrintWriter(System.out, true);

    // The Awesome Input Code is a fast IO method //
    static class AwesomeInput {
        private InputStream letsDoIT;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private SpaceCharFilter filter;

        private AwesomeInput(InputStream incoming) {
            this.letsDoIT = incoming;
        }

        public int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = letsDoIT.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        private long ForLong() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));
            return res * sgn;
        }

        private String ForString() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            }
            while (!isSpaceChar(c));

            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if (filter != null)
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            boolean isSpaceChar(int ch);
        }
    }

    // functions to take input//
    static int ni() {
        return (int) input.ForLong();
    }

    static String ns() {
        return input.ForString();
    }

    static long nl() {
        return input.ForLong();
    }

    static double nd() throws IOException {
        return Double.parseDouble(new BufferedReader(new InputStreamReader(System.in)).readLine());
    }

    //functions to give output
    static void pl() {
        pw.println();
    }

    static void p(Object o) {
        pw.print(o + " ");
    }

    static void pws(Object o) {
        pw.print(o + "");
    }

    static void pl(Object o) {
        pw.println(o);
    }

    // Fast Sort is Radix Sort
    public static int[] fastSort(int[] f) {
        int n = f.length;
        int[] to = new int[n];
        {
            int[] b = new int[65537];
            for (int i = 0; i < n; i++) b[1 + (f[i] & 0xffff)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < n; i++) to[b[f[i] & 0xffff]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        {
            int[] b = new int[65537];
            for (int i = 0; i < n; i++) b[1 + (f[i] >>> 16)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < n; i++) to[b[f[i] >>> 16]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        return f;
    }

    public static void main(String[] args) {      //threading has been used to increase the stack size.

        new Thread(null, null, "Vengeance", 1 << 25)  //the last parameter is stack size desired.
        {
            public void run() {
                try {
                    double s = System.currentTimeMillis();
                    MainSolution();
                    //pl(("\nExecution Time : " + ((double) System.currentTimeMillis() - s) / 1000) + " s");
                    pw.flush();
                    pw.close();
                } catch (Exception e) {
                    e.printStackTrace();
                    System.exit(1);
                }
            }
        }.start();
    }
}